Question

in triangle ABC, the altitude AP and BQ meet at o. show that AO* PO = BO*QO

Answer 1

Answer:

The proportionality is proved with the help of AA similarity.

Step-by-step explanation:

Given triangle ABC, the altitude AP and BQ meet at O. we have to show that $AO\times PO=BO\times QO$

In ΔAOQ and ΔBOP

∠OPB=∠OQA           (∵each 90°)

∠BOP=∠AOQ           (∵vertically opposite angles)

By AA similarity rule, ΔAOQ is similar to ΔBOP

∴ by theorem, sides are proportional as

$\frac{BO}{AO}=\frac{PO}{QO}$

⇒ $AO\times PO=BO\times QO$

Hence, proved

Answer 2

Answer:

The proportionality is proved with the help of AA similarity.

Step-by-step explanation:

Given triangle ABC, the altitude AP and BQ meet at O. we have to show that

In ΔAOQ and ΔBOP

∠OPB=∠OQA           (∵each 90°)

∠BOP=∠AOQ           (∵vertically opposite angles)

By AA similarity rule, ΔAOQ is similar to ΔBOP

∴ by theorem, sides are proportional as

⇒

Hence, proved

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